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我有一个创建发票的表单，发件人和收件人表单输入取自数据库中的可用列表。

accounts(id,...., company_name, abn) ---> id is primary
users(id, username, title, firstname, surname,...) ---> id is primary
accounts_users(id, account_id, user_id) --->account_id from accounts, user_id from users

每个用户只能属于一个帐户，但一个帐户可以有多个用户。根据登录用户的身份，表单输入是 account_id。

在发票控制器中：

 $accounts3=$this->User->AccountsUser->find('list', array(
        'fields'=>array('account_id','account_id'),'conditions' => array(
        'user_id' => $this->Auth->user('id'))));


$accounts=$this->User->Relationship->find('list', array('fields'=>array('receiver_id'),'conditions' =>  array('sender_id' => $accounts2)));

在 addInvoice 视图中：

<?php
echo $this->Form->create('Invoice', array('action'=>'addinvoice'));
echo $this->Form->input('sender_id',array('label'=>'Sender: ', 'options' => array('$accounts3' => 'COMPANY NAME')));
echo $this->Form->input('receiver_id',array('label'=>'Receiver: ', 'type' => 'select', 'options' => $accounts));
echo $this->Form->input('active', array('label' => 'Schedule?', 'options' => array('1'=>'Send','0'=>'Schedule'), 'default'=>'1'));
echo $this->Form->hidden('templates_id', array('default'=>'0'));
echo $this->Form->end('Submit');

在这种情况下，“公司名称”的标签是硬编码的，需要用表中的动态标签company_name替换accounts。

更复杂一点，接收者标签需要company_name来自accounts，或者如果company_name是NULL，（所以他们是个人帐户，而不是企业帐户）它将username来自users。

尝试的解决方案：***$sendername在InvoicesController

$sendername=$this->User->Account->find('list', array('fields'=>array('company_name'),'conditions' => array('account_id' => $accounts3)));

或者

$sendername=$this->User->AccountsUser->find('list', array('fields'=>array('id','company_name'),'conditions' =>  array('user_id' => $this->Auth->user('id'))));

这些似乎不起作用，想知道在 CakePHP 中是否有可能？

4

1 回答 1

0 
$accounts3 = $this->User->AccountsUser->Account->find('list', array(
    'fields' => array('company_name'),
    'joins' => array(
        array(
            'alias' => 'AccountsUser',
            'table' => 'accounts_users',
            'type' => 'left',
            'conditions' => array('AccountsUser.account_id = Account.id'),
        )    
    ),
    'conditions' => array(
        'AccountsUser.user_id' => $this->Auth->user('id'),
    )
));

// find all accounts with id => label from accounts table
// where AccountsUser.user_id = id of logged in user
// label being user.username when company_name is null or company_null if not null
$this->User->AccountsUser->Account->virtualFields['label'] = 'CASE WHEN Account.company_name IS NULL THEN User.username ELSE Account.company_name END';
$accounts = $this->User->AccountsUser->Account->find('list', array(
    'fields' => array('label'),
    'joins' => array(
        array(
            'alias' => 'AccountsUser',
            'table' => 'accounts_users',
            'type' => 'left',
            'conditions' => array('AccountsUser.account_id = Account.id'),
        ),
        array(
            'alias' => 'User',
            'table' => 'users',
            'type' => 'left',
            'conditions' => array('AccountsUser.user_id = User.id'),
        ),    
    ),
    'conditions' => array(
        'AccountsUser.user_id' => $this->Auth->user('id'),
    )
));

不确定这是否是您想要的，以及我是否正确理解了这个问题，但希望它能给您一些启发。

于 2012-08-07T19:52:52.270 回答